- Email: [email protected]
On upper bound for the quantum entropy
Linear Algebra and its Applications 329 (2001) 89–96 www.elsevier.com/locate/laa
On upper bound for the quantum entropy W. Hebisch a,b , R. Olkiewicz c,∗ , B. Zegarlinski d a Institute of Mathematics, University of Wrocław, Wrocław, Poland b Institute of Mathematics, Polish Academy of Sciences, Wrocław, Poland c Institute of Theoretical Physics, University of Wrocław, PL-Maxa Borna 9, 50-204 Wrocław, Poland d Department of Mathematics, Imperial College, London, UK
Received 31 July 2000; accepted 1 December 2000 Submitted by R. Bhatia
Abstract A new upper bound for the von Neumann entropy of a state of a compound quantum system is given. This leads to a log-Sobolev inequality on the matrix algebra. © 2001 Elsevier Science Inc. All rights reserved. Keywords: Logarithmic entropy; Log-Sobolev inequality
1. Introduction Quantum entropy or the von Neumann entropy of a density matrix is a concept which appears in several fields and it is in the center of interest both for mathematicians and theoretical physicists. The quantum entropy of a state describing a physical system expresses the uncertainty or randomness of that system. Similarly, the relative entropy is an information measure representing the uncertainty of a state with respect to another state. Since in statistical mechanics only a few models can be solved explicitly, the search for two-sided estimations of thermodynamical variables continues. For example, lower and upper bounds for the relative entropy proved to be useful in obtaining bounds on the free energy [5]. It was shown there that 1 1 Tr A − A1−p B p Tr A(log A − log B) Tr A1+p B −p − A (1) p p ∗ Corresponding author. Fax: +48-71-3214454.
E-mail address: [email protected] (R. Olkiewicz). 0024-3795/01/$ - see front matter 2001 Elsevier Science Inc. All rights reserved. PII: S 0 0 2 4 - 3 7 9 5 ( 0 1 ) 0 0 2 4 4 - 0
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W. Hebisch et al. / Linear Algebra and its Applications 329 (2001) 89–96
for 0 < p 1 and density matrices A, B. This inequality was next generalized in [3] to the following one: 1 Tr A log B −p/2 Ap B −p/2 Tr A(log A − log B) p 1 Tr A log Ap/2 B −p Ap/2 p for all p > 0 and positive definite A, B. In particular, it implies that 1 Tr A(log A − log B) Tr A1+p B −p − A p holds for any p > 0. In the special case of B = (Tr A)I , we have that A 1 Tr A1+p Tr A log − Tr A . Tr A p (Tr A)p
(2)
Another upper bound for the relative entropy of a state A with respect to a completely mixed state comes from log-Sobolev inequalities. Such inequalities lead to the hypercontractivity property of the corresponding semigroup [4], and hence they allow us to determine long time behavior of the quantum system. For example, it was shown in [1] that for a positive semi-definite m × m matrix A there is an optimal constant depending only on the dimension m such that 2 A a(m) tr A − tr A1/2 (3) tr A log tr A with m log(m − 1), a(2) = 2. (4) a(m) = m−2 Here tr denotes the normalized trace on m × m complex matrices. In fact it was shown for a discrete m-point space, but that directly implies (3). However, such an inequality is not useful from the physical point of view because the operator determined by the quadratic form tr(f − tr f )2 , f = A1/2 , of the right-hand side of (3) is nonlocal. It would be much more interesting to replace it by a local one, given by averages over finite subsystems. The objective of this paper is to provide such an inequality for the matrix algebra being the composition of algebras of identical finite quantum systems.
2. Main result Throughout this paper Mk×k denotes the algebra of k × k complex matrices and Tr stands for the usual trace. By tr we denote the normalized trace, that is tr(I ) = 1, where I is the identity matrix. If A = A∗ , then we call A Hermitian and A 0 means that A is positive semi-definite. For any p 1 we define the norm · p on Mk×k by
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Ap = (tr|A|p )1/p , where |A| denotes the absolute value of matrix A. By · ∞ we denote the usual operator norm. Finally, let us recall that a tr-symmetric semigroup Pt : Mk×k → Mk×k , t 0, of linear operators is hypercontractive if it is contractive in · 1 norm, and there is T4,2 > 0 such that for any t > T4,2 and all A ∈ Mk×k we have Pt A4 A2 . A semigroup Pt is tr-symmetric if tr APt B = tr(Pt A)B for any A, B ∈ Mk×k . General principles, namely, interpolation and duality, imply that Pt is then a contraction in all · p norms, p ∈ [1, ∞], and that it is also contractive from · q to · p norms, for any 1 < q < p < ∞, if t is sufficiently large. Let A = Mk×k , k 2, and A(n) = ⊗n A, the tensor product of n-copies of A. Hence, A(n) = Mk n ×k n . A(n) , as a Hilbert space with the scalar product given by f, g = Tr f ∗ g, can be decomposed as A(n) = nm=0 Hm , where Hm is a subspace connected with the range of a projector trXc (id − trj ). |X|=m
j ∈X
Here X is a subset of the set {1, 2, . . . , n}, Xc stands for its complement and |X| denotes the number of its elements. For any X = {i1 , . . . , ik }, trX denotes the conditional expectation given by trX = tri1 ◦ tri2 ◦ · · · ◦ trik , where tri is the projector from A(n) onto A ⊗ · · · ⊗ A ⊗ I ⊗ A ⊗ · · · ⊗ A, the identity matrix on the ith position, that is, tri is the partial trace with respect to the ith variable. In particular, tr = tr12,...,n . Hence, (iˆ = {i}c ) f ∈ H0 ⇔ tr12,...,n f = f, n triˆ f = f, f ∈ H1 ⇔ i=1
.. . f ∈ Hn−1 ⇔ f ∈ Hn ⇔
n i=1 n
tri f = f, tri f = 0.
i=1
˜ = {A ∈ Alternative description of the subspaces Hm can be given as follows. Let A ˜ A: tr A = 0}. Then A = A ⊕ CI . Let I = (i1 , . . . , in ) with ij ∈ {0, 1}. We put |I | = nk=1 ik . Clearly, 0 |I | n. Now let us define
if ik = 0, ˜ ik = I A ˜ A if ik = 1, ˜I =A ˜ i1 ⊗ · · · ⊗ A ˜ in . Then we have and A
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f ∈ Hm ⇔ f =
fI ,
˜ I. fI ∈ A
|I |=m
Moreover, such a decomposition is unique. In this section, we show that for any Hermitian f ∈ A(n) the following log-Sobolev inequality holds: tr f 2 log f 2 − tr f 2 log tr f 2 c(k)
n
tr(f − tri f )2 .
(5)
i=1
It is worth noting that the right-hand side of (5) can be written as f, Lf , where L = i (id − tri ). It generates a tr-symmetric semigroup Pt = exp(−tL) on A(n) . Clearly, Pt f = e−mt f for any f ∈ Hm . Let us start with the following observation. Lemma 2.1. There is a b > 0 such that f 4 b m f 2 for any f = f ∗ ∈ Hm . Proof. Because f is Hermitian, so in the decomposition f = |I |=m fI each fI are Hermitian too. Step 1: At first let us notice that A4 k 1/4A2 for any A ∈ A. Hence, for a ˜ I we have homogeneous element fI ∈ A fI 4 k m/4 fI 2 .
(6)
Moreover, by Hölder inequality, |tr f1 f2 f3 f4 | f1 4 f2 4 f3 4 f4 4 .
(7)
So f 44 =tr
4 fI
|I |=m
= = =
I,J,K,M I,J,K,M
tr fI fJ fK fM tr fI fJ fK fM k m fI 2 fJ 2 fK 2 fM 2 .
(8)
I,J,K,M
The symbol denotes the summation over all multi-indices I = {i1 , . . . , in }, J = {j1 , . . . , jn }, K = {k1 , . . . , kn }, M = {m1 , . . . , mn }, such that |I | = |J | = |K| = |M| = m and il + jl + kl + ml = / 1 for every l ∈ {1, . . . , n}. On the other hand tr fI fJ = fI 22 , f 22 = I,J
|I |=m
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93
because I = / J ⇒ tr fI fJ = 0. Step 2: Let D be a space of diagonal 3 × 3 matrices. The tensor product D (n) = (n) may be thought ⊗n D has a similar decomposition D (n) = nm=0 HD m . Clearly, D of as an algebra of functions on an appropriate discrete space. Hence, by (3) and using the tensor product property of log-Sobolev inequalities on function spaces (see, for example, Theorem 2.3 in [2]) we obtain, for any Hermitian g ∈ D (n) , the following inequality: tr g 2 log g 2 − tr g 2 log tr g 2 cD
n
tr (g − tri g)2 ,
i=1
with optimal constant cD = a(3) = 3 log 2 (see formula (4)). It implies that D = semigroup Pt , when reduced to D (n) , is hypercontractive and so for t T4,2 D (1/4)cD log 3 we have Pt g4 g2 . The formula T4,2 = (1/4)cD log 3 follows from the fact that log-Sobolev inequality implies directly that d log Pt gq(t ) 0, dt D ) = 4 so the forwhere q(t) = 1 + e4t /cD , and so Pt gq(t ) g2 . Because q(T4,2 mula follows. Suppose now that g ∈ HD m . Then
PT D g4 = e−T4,2 m g4 g2 , D
4,2
D such that and so there is a constant C = exp T4,2
g4 C m g2 .
(9)
Suppose further that g = |I |=m gI , where gI ∈ D˜ I and is given by gI = fI 2 ⊗nl=1 d il , d = diag(21/2, −2−1/2 , −2−1/2). By definition d 0 = diag(1, 1, 1). Then gI 22 = fI 22 = f 22 . (10) g22 = |I |=m
|I |=m
On the other hand g44 =
tr gI gJ gK gM
I,J,K,M
=
I,J,K,M
fI 2 fJ 2 fK 2 fM 2
n
tr θl ,
l=1
0 = 1, tr d = 0, tr d 2 = 1, tr d 3 = 2−1/2 and where θl = d il +jl +kl +ml . Because tr d 4 tr d = 3/2 so, firstly, we can replace by as in step 1, and, secondly, we can / d for all l) estimate (for all those I, J, K, M such that θl =
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p tr θl 2−1/2 ,
l=1
where p is the largest possible number of all indices l for which il + jl + kl + ml = 3. where [·] denotes the integer part of a number. Hence, Clearly, p−m [4m/3], 2/3 , and so tr θ δ , δ = 2 l l
g44 δ −m
fI 2 fJ 2 fK 2 fM 2 .
(11)
I,J,K,M
Step 3: By (8)–(11) we have f 44 k m
fI 2 fJ 2 fK 2 fM 2
I,J,K,M
k m δ m g44 k m δ m C 4m g42 k m δ m C 4m f 42 , and so f 4 bm f 2 for b = 21/6k 1/4 C.
Using the above result we now show our main inequality. Theorem 2.2. Suppose that c(k) = (2/3) log 2 + 3 log 2 log 3 + log k, and let f = f ∗ ∈ A(n) . Then inequality (5) holds. Proof. At first we show that for t > T4,2 , T4,2 = log b, where b is the constant from the above lemma, we have Pt f 4 f 2 for f = f ∗ . Let πm denote the = log b + &, & > 0. Then for t T orthogonal projector onto Hm . Suppose T4,2 4,2 n Pt f 4 = Pt πm f m=0
n m=0 n
4
e−t m πm f 4 −t m e b πm f 2
m=0
Mf 2 , where M = 1/(1 − q) and q = e−t b < 1. We now consider f ⊗ f ∈ A(2n) . By the same argument we obtain that Pt ⊗ Pt (f ⊗ f )4 Mf ⊗ f 2 , and so Pt f 4 M 1/2f 2 . Hence, by induction, Pt f 4 f 2 . But & was arbitrary, so Pt f 4
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f 2 for any t > T4,2 = log b. Finally, we take c(k) = 4T4,2 . In the following, in order to simplify notation, we denote T4,2 by T . Let z = t + iy, where t ∈ [0, T ] and y ∈ R. Because Piy 2,2 1 and PT +iy 4,2 1 so, by the Stein interpolation theorem [6], we obtain that Pt s(t ) 1, where t −1 T −t 4T + . s(t) = = 2T 4T 2T − t Suppose now that q(t) = 1 + e4t /c(k). Then q(t) s(t) for any t ∈ [0, T ], and so Pt q(t ),2 1. It implies that log Pt f q(t ) log f q(0) . Therefore, by taking the right-hand side derivative of the function t → log Pt f q(t ) at t = 0, which is also not grater than 0, the inequality tr f 2 log f 2 − tr f 2 log tr f 2 c(k)
n
tr(f − tri f )2
i=1
follows.
3. Conjecture Using properties of hypercontractive semigroups we have shown that inequality (5) holds true. However, the constant c(k) is not optimal, and we cannot get such in this way. For n = 2 we show that this constant may be reduced to a(k) = (k/(k − 2)) log(k − 1). Proposition 3.1. Suppose f = f ∗ ∈ A(2) . Then f2 a(k) tr(f − tri f )2 . 2 tr f 2
tr f 2 log
(12)
i=1
Proof. First observe that, by (3) and the tensor product property of log-Sobolev inequality on function spaces, (12) holds for all diagonal matrices f ∈ A(2) . Suppose now that f is an arbitrary Hermitian matrix. Clearly, without any loss of generality, we may assume that f is normalized, that is, tr f 2 = 1. Then inequality (12) can be written as 1 tr f 2 log f 2 , h(f ) 2 − (13) a(k) where the function h is defined by h(f ) = tr(tr1 f )2 + tr(tr2 f )2 . Let us notice that h is invariant with respect to a unitary transformation f → U1 ⊗ U2 · f · U1∗ ⊗ U2∗ , where U matrices from A. Therefore, we may assume that 1 and U2 are unitary tr1 f = ki=1 bi Pi and tr2 f = ki=1 ci Qi (we identify here A ⊗ I and I ⊗ A with
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W. Hebisch et al. / Linear Algebra and its Applications 329 (2001) 89–96
A), where {Pi } and {Qi } are sets of one-dimensional orthogonal projectors in A. Then h(f ) = (1/k) i (bi2 + ci2 ). Because f is Hermitian, so there exists a unitary matrix in A(2) such that f0 = Uf U ∗ = aij Pi ⊗ Qj . i,j
Since inequality (13) is true for diagonal matrices and the right-hand side of it is unitary invariant, so it suffices to show that h(U ∗ f0 U ) h(U0∗ f0 U0 ) for some unitary U0 being a permutation of a set of k 2 -elements. Because (f )ij,ij = (U ∗ f0 U )ij,ij = alm |Ulm,ij |2 , l,m
so bi =
1 alm |Ulm,ij |2 , k
cj =
j,l,m
1 alm |Ulm,ij |2 . k i,l,m
|2 .
Let us define γlm,ij = |Ulm,ij It is a × k 2 symmetric, stochastically positive matrix. Because h(f ) is convex with respect to γlm,ij , so it suffices to consider only extreme points of the set of all such γ . However, it is well known that this set consists of permutation matrices of a set of k 2 -elements. k2
Computer simulations shows that inequality (5) may hold with the constant a(k) also for n > 2. Therefore, we propose the following conjecture: the optimal constant copt(k) in inequality (5) equals to (k/(k − 2)) log(k − 1).
References [1] P. Diaconis, L. Saloff-Coste, Logarithmic Sobolev inequalities for finite Markov chains, Ann. Appl. Prob. 6 (1996) 695–750. [2] L. Gross, Logarithmic Sobolev inequalities and contractivity properties of semigroups, in: Lecture Notes in Mathematics, vol. 1563, 1993, Springer, Berlin. [3] F. Hiai, D. Petz, The Golden–Thompson trace inequality is complemented, Linear Algebra Appl. 181 (1993) 153–185. [4] R. Olkiewicz, B. Zegarlinski, Hypercontractivity in noncommutative Lp spaces, J. Funct. Anal. 161 (1999) 246–285. [5] M.B. Ruskai, F.H. Stillinger, Convexity inequalities for estimating free energy and relative entropy, J. Phys. A 23 (1990) 2421–2437. [6] E.M. Stein, Interpolation of linear operators, Trans. Amer. Math. Soc. 83 (1956) 482–492.
On upper bound for the quantum entropy W. Hebisch a,b , R. Olkiewicz c,∗ , B. Zegarlinski d a Institute of Mathematics, University of Wrocław, Wrocław, Poland b Institute of Mathematics, Polish Academy of Sciences, Wrocław, Poland c Institute of Theoretical Physics, University of Wrocław, PL-Maxa Borna 9, 50-204 Wrocław, Poland d Department of Mathematics, Imperial College, London, UK
Received 31 July 2000; accepted 1 December 2000 Submitted by R. Bhatia
Abstract A new upper bound for the von Neumann entropy of a state of a compound quantum system is given. This leads to a log-Sobolev inequality on the matrix algebra. © 2001 Elsevier Science Inc. All rights reserved. Keywords: Logarithmic entropy; Log-Sobolev inequality
1. Introduction Quantum entropy or the von Neumann entropy of a density matrix is a concept which appears in several fields and it is in the center of interest both for mathematicians and theoretical physicists. The quantum entropy of a state describing a physical system expresses the uncertainty or randomness of that system. Similarly, the relative entropy is an information measure representing the uncertainty of a state with respect to another state. Since in statistical mechanics only a few models can be solved explicitly, the search for two-sided estimations of thermodynamical variables continues. For example, lower and upper bounds for the relative entropy proved to be useful in obtaining bounds on the free energy [5]. It was shown there that 1 1 Tr A − A1−p B p Tr A(log A − log B) Tr A1+p B −p − A (1) p p ∗ Corresponding author. Fax: +48-71-3214454.
E-mail address: [email protected] (R. Olkiewicz). 0024-3795/01/$ - see front matter 2001 Elsevier Science Inc. All rights reserved. PII: S 0 0 2 4 - 3 7 9 5 ( 0 1 ) 0 0 2 4 4 - 0
90
W. Hebisch et al. / Linear Algebra and its Applications 329 (2001) 89–96
for 0 < p 1 and density matrices A, B. This inequality was next generalized in [3] to the following one: 1 Tr A log B −p/2 Ap B −p/2 Tr A(log A − log B) p 1 Tr A log Ap/2 B −p Ap/2 p for all p > 0 and positive definite A, B. In particular, it implies that 1 Tr A(log A − log B) Tr A1+p B −p − A p holds for any p > 0. In the special case of B = (Tr A)I , we have that A 1 Tr A1+p Tr A log − Tr A . Tr A p (Tr A)p
(2)
Another upper bound for the relative entropy of a state A with respect to a completely mixed state comes from log-Sobolev inequalities. Such inequalities lead to the hypercontractivity property of the corresponding semigroup [4], and hence they allow us to determine long time behavior of the quantum system. For example, it was shown in [1] that for a positive semi-definite m × m matrix A there is an optimal constant depending only on the dimension m such that 2 A a(m) tr A − tr A1/2 (3) tr A log tr A with m log(m − 1), a(2) = 2. (4) a(m) = m−2 Here tr denotes the normalized trace on m × m complex matrices. In fact it was shown for a discrete m-point space, but that directly implies (3). However, such an inequality is not useful from the physical point of view because the operator determined by the quadratic form tr(f − tr f )2 , f = A1/2 , of the right-hand side of (3) is nonlocal. It would be much more interesting to replace it by a local one, given by averages over finite subsystems. The objective of this paper is to provide such an inequality for the matrix algebra being the composition of algebras of identical finite quantum systems.
2. Main result Throughout this paper Mk×k denotes the algebra of k × k complex matrices and Tr stands for the usual trace. By tr we denote the normalized trace, that is tr(I ) = 1, where I is the identity matrix. If A = A∗ , then we call A Hermitian and A 0 means that A is positive semi-definite. For any p 1 we define the norm · p on Mk×k by
W. Hebisch et al. / Linear Algebra and its Applications 329 (2001) 89–96
91
Ap = (tr|A|p )1/p , where |A| denotes the absolute value of matrix A. By · ∞ we denote the usual operator norm. Finally, let us recall that a tr-symmetric semigroup Pt : Mk×k → Mk×k , t 0, of linear operators is hypercontractive if it is contractive in · 1 norm, and there is T4,2 > 0 such that for any t > T4,2 and all A ∈ Mk×k we have Pt A4 A2 . A semigroup Pt is tr-symmetric if tr APt B = tr(Pt A)B for any A, B ∈ Mk×k . General principles, namely, interpolation and duality, imply that Pt is then a contraction in all · p norms, p ∈ [1, ∞], and that it is also contractive from · q to · p norms, for any 1 < q < p < ∞, if t is sufficiently large. Let A = Mk×k , k 2, and A(n) = ⊗n A, the tensor product of n-copies of A. Hence, A(n) = Mk n ×k n . A(n) , as a Hilbert space with the scalar product given by f, g = Tr f ∗ g, can be decomposed as A(n) = nm=0 Hm , where Hm is a subspace connected with the range of a projector trXc (id − trj ). |X|=m
j ∈X
Here X is a subset of the set {1, 2, . . . , n}, Xc stands for its complement and |X| denotes the number of its elements. For any X = {i1 , . . . , ik }, trX denotes the conditional expectation given by trX = tri1 ◦ tri2 ◦ · · · ◦ trik , where tri is the projector from A(n) onto A ⊗ · · · ⊗ A ⊗ I ⊗ A ⊗ · · · ⊗ A, the identity matrix on the ith position, that is, tri is the partial trace with respect to the ith variable. In particular, tr = tr12,...,n . Hence, (iˆ = {i}c ) f ∈ H0 ⇔ tr12,...,n f = f, n triˆ f = f, f ∈ H1 ⇔ i=1
.. . f ∈ Hn−1 ⇔ f ∈ Hn ⇔
n i=1 n
tri f = f, tri f = 0.
i=1
˜ = {A ∈ Alternative description of the subspaces Hm can be given as follows. Let A ˜ A: tr A = 0}. Then A = A ⊕ CI . Let I = (i1 , . . . , in ) with ij ∈ {0, 1}. We put |I | = nk=1 ik . Clearly, 0 |I | n. Now let us define
if ik = 0, ˜ ik = I A ˜ A if ik = 1, ˜I =A ˜ i1 ⊗ · · · ⊗ A ˜ in . Then we have and A
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f ∈ Hm ⇔ f =
fI ,
˜ I. fI ∈ A
|I |=m
Moreover, such a decomposition is unique. In this section, we show that for any Hermitian f ∈ A(n) the following log-Sobolev inequality holds: tr f 2 log f 2 − tr f 2 log tr f 2 c(k)
n
tr(f − tri f )2 .
(5)
i=1
It is worth noting that the right-hand side of (5) can be written as f, Lf , where L = i (id − tri ). It generates a tr-symmetric semigroup Pt = exp(−tL) on A(n) . Clearly, Pt f = e−mt f for any f ∈ Hm . Let us start with the following observation. Lemma 2.1. There is a b > 0 such that f 4 b m f 2 for any f = f ∗ ∈ Hm . Proof. Because f is Hermitian, so in the decomposition f = |I |=m fI each fI are Hermitian too. Step 1: At first let us notice that A4 k 1/4A2 for any A ∈ A. Hence, for a ˜ I we have homogeneous element fI ∈ A fI 4 k m/4 fI 2 .
(6)
Moreover, by Hölder inequality, |tr f1 f2 f3 f4 | f1 4 f2 4 f3 4 f4 4 .
(7)
So f 44 =tr
4 fI
|I |=m
= = =
I,J,K,M I,J,K,M
tr fI fJ fK fM tr fI fJ fK fM k m fI 2 fJ 2 fK 2 fM 2 .
(8)
I,J,K,M
The symbol denotes the summation over all multi-indices I = {i1 , . . . , in }, J = {j1 , . . . , jn }, K = {k1 , . . . , kn }, M = {m1 , . . . , mn }, such that |I | = |J | = |K| = |M| = m and il + jl + kl + ml = / 1 for every l ∈ {1, . . . , n}. On the other hand tr fI fJ = fI 22 , f 22 = I,J
|I |=m
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93
because I = / J ⇒ tr fI fJ = 0. Step 2: Let D be a space of diagonal 3 × 3 matrices. The tensor product D (n) = (n) may be thought ⊗n D has a similar decomposition D (n) = nm=0 HD m . Clearly, D of as an algebra of functions on an appropriate discrete space. Hence, by (3) and using the tensor product property of log-Sobolev inequalities on function spaces (see, for example, Theorem 2.3 in [2]) we obtain, for any Hermitian g ∈ D (n) , the following inequality: tr g 2 log g 2 − tr g 2 log tr g 2 cD
n
tr (g − tri g)2 ,
i=1
with optimal constant cD = a(3) = 3 log 2 (see formula (4)). It implies that D = semigroup Pt , when reduced to D (n) , is hypercontractive and so for t T4,2 D (1/4)cD log 3 we have Pt g4 g2 . The formula T4,2 = (1/4)cD log 3 follows from the fact that log-Sobolev inequality implies directly that d log Pt gq(t ) 0, dt D ) = 4 so the forwhere q(t) = 1 + e4t /cD , and so Pt gq(t ) g2 . Because q(T4,2 mula follows. Suppose now that g ∈ HD m . Then
PT D g4 = e−T4,2 m g4 g2 , D
4,2
D such that and so there is a constant C = exp T4,2
g4 C m g2 .
(9)
Suppose further that g = |I |=m gI , where gI ∈ D˜ I and is given by gI = fI 2 ⊗nl=1 d il , d = diag(21/2, −2−1/2 , −2−1/2). By definition d 0 = diag(1, 1, 1). Then gI 22 = fI 22 = f 22 . (10) g22 = |I |=m
|I |=m
On the other hand g44 =
tr gI gJ gK gM
I,J,K,M
=
I,J,K,M
fI 2 fJ 2 fK 2 fM 2
n
tr θl ,
l=1
0 = 1, tr d = 0, tr d 2 = 1, tr d 3 = 2−1/2 and where θl = d il +jl +kl +ml . Because tr d 4 tr d = 3/2 so, firstly, we can replace by as in step 1, and, secondly, we can / d for all l) estimate (for all those I, J, K, M such that θl =
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W. Hebisch et al. / Linear Algebra and its Applications 329 (2001) 89–96 n
p tr θl 2−1/2 ,
l=1
where p is the largest possible number of all indices l for which il + jl + kl + ml = 3. where [·] denotes the integer part of a number. Hence, Clearly, p−m [4m/3], 2/3 , and so tr θ δ , δ = 2 l l
g44 δ −m
fI 2 fJ 2 fK 2 fM 2 .
(11)
I,J,K,M
Step 3: By (8)–(11) we have f 44 k m
fI 2 fJ 2 fK 2 fM 2
I,J,K,M
k m δ m g44 k m δ m C 4m g42 k m δ m C 4m f 42 , and so f 4 bm f 2 for b = 21/6k 1/4 C.
Using the above result we now show our main inequality. Theorem 2.2. Suppose that c(k) = (2/3) log 2 + 3 log 2 log 3 + log k, and let f = f ∗ ∈ A(n) . Then inequality (5) holds. Proof. At first we show that for t > T4,2 , T4,2 = log b, where b is the constant from the above lemma, we have Pt f 4 f 2 for f = f ∗ . Let πm denote the = log b + &, & > 0. Then for t T orthogonal projector onto Hm . Suppose T4,2 4,2 n Pt f 4 = Pt πm f m=0
n m=0 n
4
e−t m πm f 4 −t m e b πm f 2
m=0
Mf 2 , where M = 1/(1 − q) and q = e−t b < 1. We now consider f ⊗ f ∈ A(2n) . By the same argument we obtain that Pt ⊗ Pt (f ⊗ f )4 Mf ⊗ f 2 , and so Pt f 4 M 1/2f 2 . Hence, by induction, Pt f 4 f 2 . But & was arbitrary, so Pt f 4
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95
f 2 for any t > T4,2 = log b. Finally, we take c(k) = 4T4,2 . In the following, in order to simplify notation, we denote T4,2 by T . Let z = t + iy, where t ∈ [0, T ] and y ∈ R. Because Piy 2,2 1 and PT +iy 4,2 1 so, by the Stein interpolation theorem [6], we obtain that Pt s(t ) 1, where t −1 T −t 4T + . s(t) = = 2T 4T 2T − t Suppose now that q(t) = 1 + e4t /c(k). Then q(t) s(t) for any t ∈ [0, T ], and so Pt q(t ),2 1. It implies that log Pt f q(t ) log f q(0) . Therefore, by taking the right-hand side derivative of the function t → log Pt f q(t ) at t = 0, which is also not grater than 0, the inequality tr f 2 log f 2 − tr f 2 log tr f 2 c(k)
n
tr(f − tri f )2
i=1
follows.
3. Conjecture Using properties of hypercontractive semigroups we have shown that inequality (5) holds true. However, the constant c(k) is not optimal, and we cannot get such in this way. For n = 2 we show that this constant may be reduced to a(k) = (k/(k − 2)) log(k − 1). Proposition 3.1. Suppose f = f ∗ ∈ A(2) . Then f2 a(k) tr(f − tri f )2 . 2 tr f 2
tr f 2 log
(12)
i=1
Proof. First observe that, by (3) and the tensor product property of log-Sobolev inequality on function spaces, (12) holds for all diagonal matrices f ∈ A(2) . Suppose now that f is an arbitrary Hermitian matrix. Clearly, without any loss of generality, we may assume that f is normalized, that is, tr f 2 = 1. Then inequality (12) can be written as 1 tr f 2 log f 2 , h(f ) 2 − (13) a(k) where the function h is defined by h(f ) = tr(tr1 f )2 + tr(tr2 f )2 . Let us notice that h is invariant with respect to a unitary transformation f → U1 ⊗ U2 · f · U1∗ ⊗ U2∗ , where U matrices from A. Therefore, we may assume that 1 and U2 are unitary tr1 f = ki=1 bi Pi and tr2 f = ki=1 ci Qi (we identify here A ⊗ I and I ⊗ A with
96
W. Hebisch et al. / Linear Algebra and its Applications 329 (2001) 89–96
A), where {Pi } and {Qi } are sets of one-dimensional orthogonal projectors in A. Then h(f ) = (1/k) i (bi2 + ci2 ). Because f is Hermitian, so there exists a unitary matrix in A(2) such that f0 = Uf U ∗ = aij Pi ⊗ Qj . i,j
Since inequality (13) is true for diagonal matrices and the right-hand side of it is unitary invariant, so it suffices to show that h(U ∗ f0 U ) h(U0∗ f0 U0 ) for some unitary U0 being a permutation of a set of k 2 -elements. Because (f )ij,ij = (U ∗ f0 U )ij,ij = alm |Ulm,ij |2 , l,m
so bi =
1 alm |Ulm,ij |2 , k
cj =
j,l,m
1 alm |Ulm,ij |2 . k i,l,m
|2 .
Let us define γlm,ij = |Ulm,ij It is a × k 2 symmetric, stochastically positive matrix. Because h(f ) is convex with respect to γlm,ij , so it suffices to consider only extreme points of the set of all such γ . However, it is well known that this set consists of permutation matrices of a set of k 2 -elements. k2
Computer simulations shows that inequality (5) may hold with the constant a(k) also for n > 2. Therefore, we propose the following conjecture: the optimal constant copt(k) in inequality (5) equals to (k/(k − 2)) log(k − 1).
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